In the feedback chapater of the Razavi book, he introduces another way to calculate the DC transfer gain of amplifiers in feedback loop.
This is given by \$\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} = \frac{A_0}{1 + \beta \cdot A_0}\$. If \$\beta \cdot A_0 \gg 1\$, then \$\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} \approx \frac{1}{\beta}\$. In the circuit below, if \$R_1\$ and \$R_2\$ are very large. Therefore, we do not affect the open-loop performance of the circuit. \$R_1\$ and \$R_2\$ form a voltage divider at the output. The feedback voltage is the voltage drop across \$R_2\$, which is \$\frac{R_2}{R_1 + R_2} V_{\mathrm{out}}\$. The feedback factor \$\beta\$ is therefore \$\frac{R_2}{R_1 + R_2}\$. The closed-loop DC voltage gain is \$\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} = 1 + \frac{R_1}{R_2}\$.
The transfer function of the inverting amplifer is well-known. It is \$\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} = – \frac{R_1}{R_2}\$. This means that its feedback factor is \$\frac{R_2}{R_1}\$. How can we apply the method I used for the derivation of the feedback factor in the non-inverting amplifier to the case of the inverting amplifier?
Best Answer
Here is the derivation for the inverting case. $$\begin{align*} V_\text{out} &= A_{0}(V_{+} - V_-) \\\\ V_\text{out} &= -A_0V_x \\\\ V_x &= V_\text{in}\frac{R_1}{R_1+R_2} + V_\text{out} \underbrace{\frac{R_2}{R_1+R_2}}_{\beta} \\\\ V_\text{out} &= -A_0V_\text{in}\frac{R_1}{R_1+R_2} - A_0V_\text{out} \underbrace{\frac{R_2}{R_1+R_2}}_{\beta} \\\\ V_\text{out} \bigg(1+A_0 \frac{R_2}{R_1+R_2} \bigg) &= -A_0V_\text{in}\frac{R_1}{R_1+R_2} \\\\ \frac{V_\text{out}}{V_\text{in}} &= - \frac{A_0 \frac{R_1}{R_1+R_2}}{1+A_0 \frac{R_2}{R_1+R_2}} \\\\ \frac{V_\text{out}}{V_\text{in}} &= - \frac{A_0R_1}{R_1+R_2+A_0R_2} \to -\frac{R_1}{R_2} \ \ \ \text{for} \ \ \ A_0 \to \infty \end{align*}$$
The associated block diagram is: -